I was studying identical particles in Quantum Mechanics, when I came across the notion of the 'exchange operator' acting on a two-particle wavefunction, $\psi_(x_1, x_2)$, in one dimension: $$ P_{12}\,\psi(x_1, x_2) = \psi(x_2, x_1)$$ The way I understand, all that $P_{12}$ does is switch the positions of the two particles. I then read about the two eigenstates of this operator, which are 'symmetric' and 'antisymmetric' and correspond to bosons and fermions, respectively: $$\psi_S(x_1, x_2) = \frac{1}{\sqrt{2}} \left(\psi(x_1, x_2) + \psi(x_2, x_1) \right) \to \text{Bosons};$$

$$\psi_A(x_1, x_2) = \frac{1}{\sqrt{2}} \left(\psi(x_1, x_2) - \psi(x_2, x_1) \right) \to \text{Fermions}.$$

My question is, what is so special about the eigenstates of this operator and why do they correspond to particles?

I have encountered this operator-particle idea in other areas as well. In Particle Physics, certain particles are described by the charge conjugation operator, for instance, which reverses the charges of particles. The symmetric and the antisymmetric eigenfunctions of the charge conjugation operator correspond to two different kinds of particles, much like in the case of bosons and fermions.

I suppose I can generalize my question to this: Does the existence of an eigenstate imply the existence of a particle, and/or vice-versa? If yes, why -- what is so special about eigenstates in particular? What about other states? Why only the eigenstates of certain operators? Thanks for your time.

  • 1
    $\begingroup$ To answer the question we need to know what the word particle means to you. The picture one holds in one's head is sometimes (usually, actually) inadequate to model a quantum mechanical system or entity. $\endgroup$
    – garyp
    Aug 30, 2020 at 13:15
  • $\begingroup$ I think the question is in the sense of spin-statistics, i.e. why do particles must comply with either FD or BE statistics... $\endgroup$ Aug 30, 2020 at 13:20

2 Answers 2


Actually it doesn't have to be from a mathematical perspective, but this only comes up in $3$ or more particles. Let's consider the case of $3$ particles as it's enough to illustrate the point.

For permutation group $S_3$ there is a 2-dimensional representation, meaning there are states $\psi(x_1,x_2,x_3)$ and $\phi(x_1,x_2,x_3)$ for which $P_{ij}\psi(x_1,x_2,x_3)\ne \lambda \psi(x_1,x_2,x_3)$, i.e. the permutation operator acting on a state does not necessarily return a multiple of that state. The condition $P_{ij}^2=\mathbb{I}$ is still statisfied for this 2-dimensional representation even if $P_{ij}\psi(x_1,x_2,x_3)$ will in general yield a combination of the $\psi(x_1,x_2,x_3)$ and $\phi(x_1,x_2,x_3)$ states. In the (canonical) Yamanouchi basis for instance, \begin{align} P_{23}=\frac{1}{2}\begin{pmatrix} -1&\sqrt{3}\\ \sqrt{3}&1\end{pmatrix} \end{align} is clearly not diagonal but does satisfy $(P_{23})^2=\mathbb{I}$.

Kaplan in

Kaplan, I.G., 2013. The Pauli exclusion principle. Can it be proved?. Foundations of Physics, 43(10), pp.1233-1251.

argues that, if one allows such multidimensional representations, one obtains a fallacy based on the idea that "The physical picture in which adding one particle changes properties of all particles cannot correspond to a system of independent identical particles." In fact, the same author in

Kaplan, I.G., 1975. The exclusion principle and indistinguishability of identical particles in quantum mechanics. Soviet Physics Uspekhi, 18(12), p.988.

shows that the mean value of the one-particle operator $\hat f (x_j)$ depends on the particle numbering if multidimensional representations of $S_n$ are allowed. In the example of that paper averages of $\hat f(x_3)$ are not identical to those of $\hat f(x_2)$ or $\hat f(x_1)$.

What is special about the one-dimensional representations, i.e. the symmetric and antisymmetric representations, is precisely that all these average values are independent of particle labelling.

For completeness, the case of 2 particles is special because the irreducible representations of $S_2$ are both 1-dimensional, and in fact are precisely the symmetric and the antisymmetric representations, and nothing else. It is only for 3 or more particles that one can have multidimensional representations.

Thus, there is no reason to believe, based on permutation symmetry alone, that states must be eigenstates of an exchange operator. It seems that one must require additional physically relevant conditions to exclude states which are NOT eigenstates.

If you have the stomach for some more advanced mathematics, there's also a discussion of permutation symmetry requirements in

Hudson, R.L. and Moody, G.R., 1976. Locally normal symmetric states and an analogue of de Finetti's theorem. Zeitschrift für Wahrscheinlichkeitstheorie und verwandte Gebiete, 33(4), pp.343-351.


My question is, what is so special about the eigenstates of this operator and why do they correspond to particles?

In QM when identical particles are indistinguishable we require the quantum state describing the particles to remain the same under particle exchange. Since multiplying a state vector by a constant doesn't change the actual state, this is equivalent to requiring our multi-particle states to be eigenstates of the exchange operator.

About the latter part of the question, I'm not sure what you mean. What else would we be describing quantum systems of? Our quantum startes describe states of particles, and so we require our theories to match what we know about these particles and how they behave.

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    $\begingroup$ There's a distinction between "identical" and "indistinguishable". One can have identical particles in two distinct well and they would not be indistinguishable (see physics.stackexchange.com/q/377078/36194 or physics.stackexchange.com/q/37556/36194. "Identity is a necessary but not sufficient property for indistinguishable particles" See Bach, Alexander, 1988, Foundations of physics, 18(6), pp.639-649 and other papers by this author, or Kaplan, Ilya.G., 1975. Soviet Physics Uspekhi, 18(12), p.988. $\endgroup$ Aug 30, 2020 at 13:22
  • $\begingroup$ @ZeroTheHero Of course, but I'm assuming the OP isn't asking about a scenario where this isn't relevant. I have edited to be more careful though. $\endgroup$ Aug 30, 2020 at 15:00

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